最终中兴捧月—教师家访程序.doc

C语言家访问题代码# include# define M 45# define N 40# define L 130int subtract(int a,int b)int result;int var1,var2; var1=a%100; var2=b%100;result=b-a;if(var2=60) result=result+40; return(result);int change(int T,int a)int result;int var1,var2;if(a100)var1=a/100;var2=a%100;a=var1*60+var2;result=T*a;var1=result/60;var2=result%60;result=var1*100+var2;return(result);main() FILE *fd,*fs;int n,s,t;int stuN3; int disN+1N+1; int (*ps)3=stu;int i,j,m,k,p;int T;int result1,result2,var,d;int min_start_start,min_start_over,min_over_start,min_over_over;int time_over; /*recode the last visit parents over time*/ /*read the data of the student.txt file*/ fs=fopen(student.txt,rb);if(fs=NULL) puts(The file open error! );else fscanf(fs,%d,&n); for(i=0;in;i+) for(j=0;j=2;j+) fscanf(fs,%d,&stuij); fclose(fs); /*read the dat of the distance.txt file*/fd=fopen(distance.txt,rb);if(fd=NULL) puts(The file open error! );else fscanf(fd,%d,&n); for(i=0;i=n;i+) for(j=0;j=n;j+) fscanf(fd,%d,&disij); fclose(fd);printf(please input the speed:the minutes for one distance:);scanf(%d,&T); /* the journey speed:the minutes for one distance*/ for(i=1;i=n;i+)for(j=1;j=n;j+)disij=change(T,disij); /* change the distance into the require time between parents*/for(p=0;pn;p+) if(p!=0) for(i=p;i*(*(ps+i)+1)&result1=*(*(ps+i)+2) if(in-1) for(j=i;jn;j+) for(k=0;k3;k+) *(*(ps+j)+k)=*(*(ps+j+1)+k); n-; /* delete the unqualified parents*/ for(i=p;in;i+) result1=subtract(*(*(ps+i)+1),*(*(ps+i)+2); if(result1M) /*judge if the parents accords with the condition*/ if(in-1) for(j=i;jn;j+) for(k=0;k3;k+) *(*(ps+j)+k)=*(*(ps+j+1)+k); n-; /* delete the unqualified parents*/ for(i=1;in-p;i+) for(j=p;j*(*(ps+j+1)+1) for(k=0;k*(*(ps+j+1)+2) for(k=0;k3;k+) var=*(*(ps+j+1)+k); *(*(ps+j+1)+k)=*(*(ps+j)+k); *(*(ps+j)+k)=var; else if(*(*(ps+j)+2)=*(*(ps+j+1)+2) if(jn-1) for(m=j+1;mn;m+) for(k=0;k3;k+) *(*(ps+m)+k)=*(*(ps+m+1)+k); n-; /*delete parents whos time are same*/ min_start_start=*(*(ps+p)+1); /*the minimum stating time*/ min_start_over=min_over_over=*(*(ps+p)+2); m=p; for(i=p+1;in;i+) if(*(*(ps+i)+2)min_over_over) min_over_over=*(*(ps+i)+2); m=i; min_over_start=*(*(ps+m)+1); /* the minimum end time*/ if(m!=p) d=dis*(ps+p)*(ps+m); result1=subtract(min_start_start,min_over_over); result1=subtract(d,result1); result2=subtract(min_over_start,min_start_over); result2=subtract(d,result2); if(result1=L) for(k=0;k0)d=dis*(ps+p)*(ps+p-1); result1=subtract(d,*(*(ps+p)+1); if(*(*(ps+p-1)+2)result1)&(result1time_over) *(*(ps+p-1)+2)=result1; if(time_over=result1) *(*(ps+p-1)+2)=time_over; /* modify the last visit parents end time*/ if(p!=n-1) time_over=*(*(ps+p)+2); /*recode the last visit parents over time*/ *(*(ps+p)+2)=add(*(*(ps+p)+1),M); /* update the next visit parentsend time*/ printf(The max visit number is:%d,n); printf(n); printf(the visit